I can't understand this in partial fractions

[arildno]

Hmm..I just reviewed your original question:
It seems to want the general termof the expansion, rather than the specific one.

1.Have you learned about the Cauchy product of series yet?

2. Have you learned about Taylorseries, and how to compute them?

 

[dilan]

you mean like |x|<1 right?

 

[dilan]

I am sort of

Hmm..I just reviewed your original question:
It seems to want the general termof the expansion, rather than the specific one.

1.Have you learned about the Cauchy product of series yet?

2. Have you learned about Taylorseries, and how to compute them?

Sorry arildno I havn't learned any of them. Will it be very difficult for me to learn it?

 

[arildno]

What sort of class are you taking?
I'm suddenly unsure as to what method your teacher had in mind when giving that exercise.

 

[dilan]

I am just in the A/L class. Just finished my Local O/Ls. I just swiched to the english medium now. I don't know, really that teacher is too advance sort of. Even the other students also sort of get fedup of maths like. But I am trying to somehow keep up. He told us that there is a big gap between O/Ls and A/Ls. He told that it is very difficult to cover this gap. He thinks that we have done everything. I don't know whether in the London sylluba it's there. But in our O/L syllubas non of this is there that you thought above except for the equation

Sn = n/2 (a+l)
and
Sn = n/2 (2a + (n-1)d)

Well that's the way. In our country only from arround 20% - #8% like pass Maths.
I am just trying to come into the standard. Well I believe that PF was the best place for me.

 

[arildno]

Seems that you've got a tough time ahead, but the right attitude to face it then!

Now, have you learned that a function f(x) may be written in series form:
[itex]f(x)=f(0)+f'(0)x+f''(0)\frac{x^{2}}{1*2}+f'''(0)\frac{x^{3}}{1*2*3}+++[/tex]<br /> where, say, f''(0) means the 2nd derivative of f, evaluated at x=0?[/itex]

 

[dilan]

Well

Seems that you've got a tough time ahead, but the right attitude to face it then!

Now, have you learned that a function f(x) may be written in series form:
[itex]f(x)=f(0)+f'(0)x+f''(0)\frac{x^{2}}{1*2}+f'''(0)\frac{x^{3}}{1*2*3}+++[/tex]<br /> where, say, f''(0) means the 2nd derivative of f, evaluated at x=0?[/itex]

$<br /> I am sort of like stuck. But I have to face these. TO be honest arildno I don't know that also. <img src="/styles/physicsforums/xenforo/smilies/cry.png" class="smilie" loading="lazy" alt=":cry:" title="Cry :cry:" data-shortname=":cry:" /> <br /> What am I to do? have any website that I can learn all these and come back to you?$

 

[arildno]

Well, it isn't as bad as you think, because what I've tried to find out was if there was an ALTERNATIVE method of solving your problem than the one I initially thought of, and that you knew about.
That is; It isn't loads and loads of new stuff you have to learn in order to tackle your original question, from what we've covered in this thread, we may well find a way to answer your question without using the concept of derivatives, Taylor series and so on.

Just give me a bit of time, OK?

 

[dilan]

Ok no problem. Realy appreciate your interest in helping me. Can't believe, but really you are great person.
Thanks a lot
Hope I will find a solution.

 

[dilan]

hi arildno,

Uhh were you able to get anything? Well it's ok take your time, I am not in a hurry.

 

[arildno]

Okay, I'll post something on the Cauchy product of infinite series first:
Suppose you've got two series $S_{1}=\sum_{n=0}^{\infty}a_{n}x^{n}, S_{2}=\sum_{m=0}^{\infty}b_{m}x^{m}$
where $a_{n},b_{m}$ are coefficients for the n'th and m'th terms respectively (it is smart, and allowable, to use a different letter for the indices for the two series)

Now, suppose we wish to multiply the two series together to a single series S; how are the coefficients of S related to the coefficients of $S_{1},S_{2}$ ?

Now, let's just use a finite amount usual numbers, and see how the result OUGHT to be:
If we are to multiply the number (2+3+1) (that is, 6) with (5+7+8) (that is 20), we may do it as follows:
$(2+3+1)(5+7+8)=2*5+2*7+2*8+2*8+3*5+3*7+3*8+1*5+1*7+1*8$
The crucial point to notice that we multiply each term in the first parenthesis with each term in the other parenthesis, and then sum all the 3*3=9 terms together (3*3 terms since each parenthesis has 3 numbers).

Thus, generalizing to an infinite series, we could write our product as:
$$S=S_{1}*S_{2}=\sum_{n=0}^{\infty}a_{n}x^{n}*\sum_{m=0}^{\infty}b_{m}x^{m}=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{n}b_{m}x^{n+m}$$
where the last DOUBLE sum should be thought of as saying:
1. Fix n=0, and calculate the sum $\sum_{m=0}^{\infty}a_{0}b_{m}x^{m+0}$
that is, multiply the term $a_{0}x^{0}$ with EACH term in $S_{2}$, and sum them together

2. ADD to your result from 1. the series you get by fixing n=1, i.e, by first calculating the sum $\sum_{m=0}^{\infty}a_{1}b_{m}x^{m+1}$, and then adding all of this to the result you got from 1.
(That is, multiplying the term $a_{1}x^{1}$ with EACH term in $S_{2}$, add up the result, and add this result with that you got from 1.)
3. And so on..

Got that?

 

[arildno]

I'll assume you do, and continue:
Now, as with any sum of numbers, it shouldn't matter in which succession you add them together, right? (That is: 1+2+3=2+1+3=3+1+2 and so on)

We wish now to add up all terms in our double sum in ASCENDING powers of x.
We introduce a new index, j=n+m
Rewriting this, we have n=j-m
Now, for FIXED j, m can have any value between 0 and j.
Thus, we get the upper and lower limits on j and m:
$0\leq{j}\leq\infty,0\leq{m}\leq{j}$
and we get, by substituting j=n+m or n=j-m:
$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{n}b_{m}x^{n+m}=\sum_{j=0}^{\infty}\sum_{m=0}^{j}a_{j-m}b_{m}x^{j}=\sum_{j=0}^{\infty}c_{j}x^{j}, c_{j}=\sum_{m=0}^{j}a_{j-m}b_{m}$$
That is, cj is the coefficient of the j'th power of x.

Example:
You were to multiply together:
$$(1+(-2x)+(-2x)^{2}+++)(1+(-2x)+(-2x)^{2}+++)$$
We can write this as the product of the series:
$$S=\sum_{n=0}^{\infty}(-2)^{n}x^{n}\sum_{m=0}^{\infty}(-2)^{m}x^{m}=\sum_{j=0}^{\infty}c_{j}x^{j}, c_{j}=\sum_{m=0}^{j}(-2)^{j-m}(-2)^{m}=\sum_{m=0}^{j}(-2)^{j}=(-2)^{j}\sum_{m=0}^{j}1=(j+1)(-2)^{j}$$

Thus, we have the series representation:
$$\frac{1}{(1+2x)^{2}}=\sum_{j=0}^{\infty}(j+1)(-2)^{j}x^{j}$$
That is, the coefficient of the j'th power of x is [itex](j+1)(-2)^{j}[/tex]<br /> Note that this is in tune for j=4 which we already have calculated:<br /> $$c_{4}=(4+1)(-2)^{4}=5*16=80$$[/itex]